3.052 SAMPLE PROBLEMS FROM PREVIOUS YEARS
1.
List 5 important scientific issues to consider when scaling down a
mechanical machine with moving components to the nanoscale.
Discuss each in a sentence or two.
ANS.
1) Friction : Nanoscale devices have large surface to volume ratios and
hence, surface effects become much more important. Factors
affecting
friction, such as surface roughness, must be controlled.
2) Adhesion : "Sticking" of surfaces together must be avoided in order for
the parts to move smoothly. This can be accomplished by reducing of
attractive surface interactions and maximizing repulsive ones and can be
addressed for example by nanoscale lubricants. Friction and adhesion are
related and combined together are sometimes called "stiction."
3) Contamination : Dust particles and other contaminants which were
negligible at the macroscale could hinder the function of the device
4) Mechanical Properties : Depending on the material chosen, the
mechanical properties at the nanoscale often will be different than at the
macroscale, e.g. anisotropy, plastic deformation, etc. and must be
considered.
5) Environment : The nanoscale environment where the device will be
used must be considered (e.g. air, water, electrolyte, etc.). If the device
is to be used inside the human body nanoscale interactions with the
biological environment must be designed in (biocompatibility). High-energy
radiation can break chemical bonds and hence, disrupt nanomachines.
6) Thermal Vibrations : At room temperature, geometrical and mechanical
property design can be used to limit thermal vibrations.
2. An Au-coated microfabricated silicon nitride (Si3N4) cantilever probe tip
(spring constant, k = 0.01 N/m, square pyramidal geometry) and planar
surfavce are functionalized with carboxylic-acid terminated short
molecules. A nanomechanical experiment is conducted in which the
sample is moved by a piezo at a constant rate towards the probe tip in
buffer solution at pH=9 conditions at which carboxylic acid groups are
ionized yielding an electrostatic double layer repulsion (see Figure 1 below).
1
10
probe tip
COOH probe tip versus COOH
surface average
D
O
electrostatic
repulsion
O
O
C
Force/Radius (mN/m)
C
O
8
0.5
COOH probe tip versus COOH
surface standard deviation
6
0.3
4
Force (nN)
cantilever
2
0.1
0
0
5
10
15
20
-2
sample
-0.1
Tip-Sample Separation Distance, D (nm)
Figure 1. Chemical force spectroscopy experiment taken using a
nanomechanical instrument called a 1-Dimensional Molecular Force Probe
(Asylum Research, Inc.); "radius" on the y-axis is the probe tip end-radius,
positive values of force indicate repulsion.
(a) At a tip-sample separation distance of 3 nm how much does the
cantilever defect upwards?
ANS. Reading off the plot above, at D = 3 nm the Force, F =0.14 nN.
Using Hooke's Law to represent the behavior of the cantilever:
F = kδ
0.14nN
F
=
= 14nm
k 0.01N / m
where is the cantilever deflection. 2 pts
(b) At D=0, estimate how many molecules are feeling the repulsive
force on the probe tip if the area per molecule, Amolecule ~ 0.216 nm2
(*hints : approximate the probe tip geometry simply and take into
account the maximum range of the repulsive interaction) 3 pts
ANS. The probe tip geometry can be approximated as a hemisphere
(Figure A1.) At D = 0 the surface area of the probe tip which feels the
repulsive force (ATIP) is shown in blue. The geometrical parameters are
defined as follows: D is the probe tip-substrate separation distance = 0,
DMAX is the maximum tip-sample separation distance range of interaction)
~ 5 nm from Figure 1, and RTIP is the probe tip radius which is determined
from the Force/Radius axis of the plot above to be = 63 nm. ATIP is the
surface area of a portion of a sphere which is defined as:
δ=
A TIP = π R TIP D MAX = π ( 63nm )(15 nm ) = 2967 nm 2
The number of molecules that can fit in this surface area is :
No. Molecules =
ATIP
Amolecule
2
2967 nm 2
=
= 13, 747
0.216 nm 2
RTIP
probe
tip
DMAX
ATIP
D=0 substrate
Figure A1.
(c) What could be the reason for the apparent infinite slope of the
curve at D=0? 2 pts
ANS. The apparently infinite slope at D=0 indicates that the surface below
the molecules is much stiffer than the transducer.
(d) Schematically draw how you would expect the force versus tipsample separation distance plot above to change if the surface was a
carboxylic acid end-functionalized polymer that is end-tethered to
the surface (Figure 2) with a height of ~ 10 nm. Indicate the correct
approximate distance range at which the forces are felt and explain
in a few sentences the molecular origin of this change. 2 pts
O
O
C
10 nm
Figure 2.
ANS. Upon approaching the surface, the electrostatic repulsion will still be
felt at 15 nm from the carboxylic acid head groups at the top of the
polymer layer. An additional 10 nm of repulsion will take place due to
entropic compression of the polymer chain. Hence, the maximum distance
range will be :
D MAX = Delectrostatic + D polymer = 15 nm + 10 nm = 25 nm
3
COOH probe tip versus COOH
surface average
8
0.5
COOH probe tip versus COOH
surface standard deviation
6
polymer surface
0.3
4
Force (nN)
Force/Radius (mN/m)
10
2
0.1
0
0
5
10
15
20
-2
-0.1
Tip-Sample Separation Distance, D (nm)
Hence, The curve will shift to the right by approximately 10 nm.
1.
Use the following paper as a reference: "Probing Intermolecular
Forces and Potentials with Magnetic Feedback Chemical Force
Microscopy," Paul D. Ashby, Liwei Chen, and Charles M. Lieber, 9467
J. Am. Chem. Soc. 2000, 122, 9467-9472 (posted on the course
website).
a. Digitize and plot the data of Figure 5 (feel free to use software
such as FindGraph, a 30 day trial version can be downloaded from
here: http://www.uniphiz.com/findgraph.htm) which is a high
resolution force spectroscopy experiment between two hydroxylterminated self-assembling monolayers.
A.
4
Digitized Data
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
Approach
Retract
-0.2
-0.4
-0.6
-0.8
-1
D (nm)
b. Fit the data to an appropriate van der Waals interaction function
(approach and retract separately) and show the fits on your plot.
A. For this part, we focus on the attractive regime of the curves - between
D=1.5 nm and D=8 nm - only. We can model the interaction between the
AFM tip and the sample surface as an interaction between a sphere and a
surface. From the Lecture #14 Supplement, the van der Waals interaction
potential between a sphere and a flat surface is:
WSPH-SFC(D)=-AR/6D
where A is the Hamaker constant, R is the radius of the sphere (which is
75nm according to the Ashby paper), and D is the separation distance
between the sphere and the surface. The van der Waals interaction force
is calculated by taking the derivative of the potential with respect to D:
FSPH-SFC(D) = -dWSPH-SFC(D)/dD
FSPH-SFC(D) = -AR/6D2
One way to fit the digitized data to an appropriate van der Waals
interaction function is to re-plot the data points in the attractive regime as
F vs. R/6D2. A linear trend-line of the data plotted in this manner will have
a slope equal to the Hamaker constant.
5
Attraction Regime Only
0
0
1
2
3
4
5
6
-0.1
-0.2
Approach
-0.3
y = -0.0976x - 0.022
-0.4
Retract
Linear
(Retract)
Linear
(Approach)
-0.5
-0.6
-0.7
-0.8
-0.9
y = -0.1782x + 0.0238
-1
R/6D^2 (1/nm)
c. From these datafits, estimate the Hamaker constant for the
approach and retract curves separately.
A. Reading off the slopes of the formulas for the above trend-lines:
Approach: A = 0.0976 nN*nm = 9.76 x 10-20 J
Retract: A = 0.1782 nN*nm = 1.782 x 10-19 J
d. How do these values compare to typical values for metals,
ceramics, and polymers?
A. From the Lecture #14, Supplement #2:
Metals
3.0-5.0 x 10-19
Alumina
1.40 x 10-19
MICA
1.35 x 10-19
PVC
7.8 x 10-20
Hydrocarbons 5.0 x 10-20
2.
The data of Fig. 13.13 in Israelachvili is a surface force apparatus
experiment between two lipid monolayers. At higher temperatures,
the interaction is suggested to be primarily dominated by
hydrophobic interactions. Digitize or manually measure this data,
plot, and fit to an appropriate van der Waals function to calculate a
Hamaker constant. Does the Hamaker constant value support this
hypothesis?
Problem 2.
Digitize the data at 37°C , as shown below, interaction force between the
two plates is zero at long distance, thus, the force curve need to be shifted to zero the
force at longer distance ( > 6nm in this case).
6
0.4
0.2
Raw Data
F/R (mN·m-1)
0
-0.2
Shifted Data
-0.4
-0.6
-0.8
-1
0
2
4
6
8
Distance (nm)
−2
Plug in the fact R = 1cm = 10 m , fit the force in attraction region with Van der waals
AR
interactions, F = −
, as shown below.
6D 2
x=-R/6D2 (nm-2)
0
0.0002
0.0004
0.0006
0
-0.002
F(mN)
-0.004
-0.006
-0.008
F = -16.835x + 0.0004
-0.01
R
is observed, this indicates that van der Waals
6D 2
interactions is not the only dominant effect in this experiment. In addition, the linear fit
yields a Hamaker constant A = 16.8mN ⋅ nm 2 = 1.68 × 10 −20 J = 16.8 zJ , greater than the
Hamaker constant of hexadecane in water = 5zJ (Israelachvili). Hence, this experiment
suggests hydrophobic interactions are present.
No linear dependence of F and −
3.
Is it possible to measure the attractive adhesion force due to
dispersive van der Waals interactions in an atomic force
7
spectroscopy experiment conducted in vacuum or air with a Si3N4
cantilever and probe tip on a flat surface at a tip-sample separation
distance, D=10 nm? (HINT: model the AFM probe tip as a sphere with
a radius equal to the tip radius of curvature, Rtip≈50 nm and assume
the Hamaker constant, A≈10-19J).
A. To calculate the attractive adhesion force due to dispersive van der
Waals interactions, refer to page 177 of Israelachvili (or Lecture Notes
#13) where the dispersion interaction energy between a sphere and
surface is defined as follows:
WSPH-SFC(D)=-AR/6D
The dispersion interaction force is calculated by taking the derivative of
the energy with respect to D:
FSPH-SFC(D) = -dWSPH-SFC(D)/dD
FSPH-SFC(D) = -AR/6D2
Substituting in the given numerical values :
FSPH-SFC (D=10 nm) = (10-19 Nm • 50•1E-9 m) /6(10•1E-9 m)2
FSPH-SFC (D=10 nm) = 8.3 pN
From problem set #1, the force detection limit of the AFM is determined by
thermal oscillations of the force transducer,<Fm2>1/2. If we assume it
behaves as a one-dimensional, free, undamped harmonic oscillator then
<Fm2>1/2 = √(k BTk ). Since the minimum k = 0.01 N/m, <Fm2>1/2 = 6.4 pN.
Therefore, yes, it is possible since FSPH-SFC (D=10 nm) > <Fm2>1/2.
4.
Q. Short Answer Questions (*please answer with a few sentences):
(a) Two parallel, nonpolar polymers are a distance r apart, as the size
of the monomer unit increases, does the attractive dispersive van
der Waals interaction increase or decrease?
A. From p. 177 Israelachvili and the Lecture #14 Supplement,
w(r)=-3πAL/8σ2r5
where σ is a parameter which describes the size of a monomer unit.
Hence, it decreases.
(b) What is the difference between the rupture force and the
adhesion force?
A. The rupture force is the force necessary to break the interaction
between individual atoms or molecules. The adhesion force is the force
needed to break the interaction between bodies containing many atoms or
molecules, e.g. particles, surfaces, etc.
8
(c) The interaction potential between bodies is a function of what
three physical parameters?
A.
(1) geometry of the experiment
(2) size of the bodies
(2) Hamaker constant or interactomic / intermolecular potential
(d) Name five applications where interparticle interactions are critical.
A.
(1) biocompatibility of implant surfaces
(2) pharmaceutical powder blending
(3) paints
(4) AFM probe tip-surface interactions
(5) metallographic polishing fluids
1.
Q. Consider the approach of two individual phenol molecules (C6H5OH) in a
high-resolution force spectroscopy experiment conducted at room
temperature in air as shown in the Figure below. Assume the phenol
molecules interact as freely rotating permanent dipoles (dipole moment,
μ=1.5 D (Debye) where 1D=3.336*10-30 C•m) which (besides attraction)
undergo a steric, soft, power law repulsion (repulsive interaction parameter,
B=10-135 Jm12, repulsive interaction exponent, m=12).
(a) Neglecting VDW dispersion forces, Plot U(r) (kBT) and F(r) (nN) for r<1.0
B
nm on separate graphs and label ro (nm), re(nm), rs(nm), Umin(kBT), and
Frupture(nN) on the graphs.
B
spring
k=0.1 N/m
intermolecular
interaction
phenol molecule
attached to probe tip
phenol
surface
A. (a) The total intermolecular interaction energy, U(r), is a summation of the
attractive and repulsive components of the potential :
9
U(r) = Uattractive(r)+Urepulsive(r)
For freely rotating permanent dipoles, the attractive component is defined as follows
(Israelachvili, page 28) :
Uattractive(r)=-(μ12μ22/3kBT(4πεo)2)r6 (1)
B
where : μ1 and μ2 are the dipole moments of the two interacting molecules (for this
problem μ1=μ2=μ=1.5 D (Debye) where 1D=3.336*10-30 C•m), εo is the dielectric
permittivity of free space=8.854*10-12 C2 J-1m-1, and kBT is the thermal energy at room
temperature= 4.1*10-21 J.
B
The steric, soft, power law repulsion (Israelachvili, page 112) :
Urepulsive (r)=+B/rm
where : the repulsive interaction parameter is given as B=10-135 Jm12 and the repulsive
interaction exponent, m=12.
The force curve is given by :
F(r)=-dU(r)/dr
10
11
re
ro rs
1
U(r)(kBT)
0.5
Uattractive
0
0
-0.5
0.2
0.4
0.6
0.8
1
0.8
1
Utotal
-1
Umin=EB
Urepulsive
-1.5
0.05
0.04
F(r)(nN)
0.03
0.02
0.01
0
-0.01 0
0.2
0.4
0.6
-0.02
-0.03
Frupture
-0.04
-0.05
r (nm)
12
(b) The spring will exhibit a linear mechanical instability when
|dF/dD|>|k|. Superimpose the spring instability curve on F(r) to
determine the separation distance at which the jump-to-contact occurs
and the separation distance range of lost data.
0.05
0.04
spring instability curve
F(r)(nN)
0.03
0.02
0.01
0
-0.01
0
0.2
0.4
0.6
0.8
1
-0.02
-0.03
-0.04
-0.05
region of
lost data
k=0.1 N/m
r(nm)
The jump-to-contact occurs at ~ r=0.45 nm.
The region of lost data takes place from ~ r=0.29 nm to r=0.45 nm.
(c) If one was to include a power law VDW dispersion interaction in the
calculation of the total intermolecular interaction, how would the binding
energy change (interaction constant, A=10-77 Jm6, interaction exponent,
n=6)?
A. We need to add one more attraction term to equation (1) :
Uattractive(r)=-(μ12μ22/3kBT(4πεo)2)r6 -A/rn
B
13
5
U(r)(kBT)
0
0
0.2
0.4
0.6
0.8
1
-5
-10
-15
The binding energy increases to EB=13kBT.
B
2.
Q. It is often argued that the Lennard-Jones potential is applicable to
chemical bonds as well as physical bonds, where the attractive van der Waals
part remains unchanged and where only a difference in the repulsive
coefficient, B, distinguishes between the two types of interaction. If A=10-77
Jm6 and the equilibrium separation is at re=0.35 nm for physical bonding and
re=0.15 nm for chemical bonding, assume that the above stated hypothesis is
true and calculate the binding energy EB=wmin from each case and argue
whether or not the result tends to support the above view.
B
A. Given : A=10-77 Jm6, re(physical)=0.35 nm, re(chemical)=0.15 nm
re=[2B/A]1/6 (page 9, Israelachivili) ⇒ B=A[re]6/2
Physical Bonds :
B=10-77 Jm6• [0.35•1E-9 (m)] 6/2=0.919•1E-134 Jm12
Chemical Bonds :
B=10-77 Jm6• [0.15•1E-9 (m)] 6/2=5.695•1E-137 Jm12
wmin=-[A2/4B] (page 9, Israelachivili)
wmin(physical)=-[10-77(Jm6)]2/(4•0.919•1E-134 Jm12) 4.1*10-21J/kBT=0.66kBT
wmin(chemical)=-[10-77(Jm6)]2/(4•5.695•1E-137 Jm12) 4.1*10-21J/kBT=107kBT
B
Yes, since the bond strength of covalent bonds range from 100-800kBT.
B
14
3.
Short Answer Questions.
(a) Q. Why does ice float on water?
A. Water has a higher density because when ice crystallizes it packs into a more
open, H-bonded structure (see Israelachvili page 122-125).
(b) Q. What is a possible reason for the low melting point, Tm, of many
traditional polymers?
A. Tm is controlled by weak, intermolecular dispersion interactions between
neighboring polymers chains.
(c) Q. What is the primary controlling intermolecular interaction for the
following polymers in the solid state :
(1) poly(urethane)
(2) poly(vinyl chloride)
(3) poly(styrene)
A.
(1) H-bonding or dipole-dipole
(2) H-bonding or dipole-dipole
(3) dispersion
(d) A -CH3 functionalized AFM probe tip is brought into contact with a
-CH3 functionalized surface in water and the adhesion force is observed to be
much greater than that expected for purely dispersive interactions, why?
A. -CH3 groups are strongly hydrophobic and prefer to stick together.
(e) Two surfaces exhibit a strong adhesion in water but not in cyclohexane or
benzene, why?
A. The surfaces are hydrophobic in water which is polar. The effect is eliminated
when dissolved in a nonpolar solvent such as cyclohexane or benzene.
Q. The limit of force detection for a particular AFM cantilever is 10 pN. Is it possible
to measure the attractive force due to dispersive van der Waals interactions in
an atomic force spectroscopy experiment conducted in vacuum or air with a
Si3N4 cantilever and probe tip (Rtip ≈50 nm) on a flat surface at a tip-sample
separation distance, D=10 nm ? You many assume the Hamaker constant,
A≈10-19J.
A. To calculate the attractive adhesion force due to dispersive van der Waals
interactions, refer to page 177 of Israelachvili where the dispersion interaction energy
between a sphere and surface is defined as follows:
W(D)=-AR/6D
The dispersion interaction force is calculated by taking the derivative of the energy
with respect to D :
F(D)=-dW(D)/dD
15
F(D)=-AR/6D2
Substituting in the given numerical values :
F(D=10 nm)=(10-19 Nm • 50•1E-9 m) /6(10•1E-9 m)2
F(D=10 nm)=8.3 pN
No you could not measure it.
2.
Q. Derive an equation for the interaction force between a charged molecule
and a surface consisting of freely rotating dipolar molecules as a function of
separation distance, D.
A. The interaction potential between a molecule and a surface is derived on page 156
of Israelachvili (or Lecture #6) as equation (10.1) :
W(D)=[2πAρ]/[(n-2)(n-3)Dn-3]
This is assuming an intermolecular pair potential of the following form:
w(r)=-A/rn
Substituting n=4 for the charge-freely rotating dipole interaction (page 28,
Israelachvili) into W(D) one obtains :
W(D)=[2πAρ]/[(4-2)(4-3)D4-3]
W(D)=[2πAρ]/[2D]
Then, one can take the derivative of the energy with respect to D to obtain the
force :
F(D)=-dW/dD=-[2πAρ]/[2D2]
(a) Explain in words what the Hamaker constant is.
A. The Hamaker constant is the strength of the molecular level interatomic or
intermolecular interaction multiplied by the density (A=Aπ2ρ2).
(c) Define adhesion hysteresis and what causes it.
A. Adhesion hysteresis is when the adhesion force on retract is greater than the jump-tocontact force and is caused by time-dependent structural rearrangements of the surface,
polymeric diffusion across the interface, water condensation due to additional capillary
forces, etc.
16
(d) Can you ever get an attractive force between two charged surfaces (with no
other attractive interactions acting) and why?
A. Yes if the two surfaces are oppositely charged.
(e) Explain what a contact area is.
A. It is the interaction area between two surfaces under an external compressive load or
due to surface adhesion forces.
1. Q. Name three differences between Nanomechanics and Macromechanics.
A.
NANOMECHANICS
MACROMECHANICS
size scale probed~nanometer,
size scale probed~millimeter
e.g. single molecules
e.g. many molecules
force range probed ~nanonewton
force range probed ~ newton
details of chemical structure important in
details of chemical structure ignored
many cases
in many cases
experimental techniques used : atomic force
experimental techniques used : bulk
microscopy, optical tweezers, surface force
mechanical testing
apparatus
surfaces important
bulk properties important
mechanical properties size-dependent
mechanical properties independent of size
(a) Would you expect an attractive or repulsive force to be present between
two hydrophilic surfaces in aqueous solution?
A. repulsive, so that they can maximize their exposure to water (this is the
opposite of the hydrophobic attractive force)
(b) Would you expect hydrophobic forces to be present in organic solvents?
A. No! The origin of the attractive hydrophobic force is due to the avoidance of
water.
(c) The attractive van der Waals dispersion force ~α2 where : α is the
polarizability of the e- cloud. Explain this.
A. As the electron cloud is more easily distorted, the strength of the fluctuating
dipoles increases and hence, the attractive force increases.
(d) What is the difference between polar interactions and polarization
interactions?
A. Polar interactions are between an ion and a permanent dipole or two permanent
dipoles. Polarization interactions are between charge-nonpolar (induced or
instantaneous dipole) dipole-nonpolar (induced dipole).
.
17
(e) Name any three requirements for studying the elasticity of individual
macromolecules.
A. low surface grafting density, the ability to tether two ends of a chain, small probe
tip (Rtip≈RF), a soft cantilever, i.e. a cantilever spring constant comprable to the the
global chain stiffess (kcantilever ≈kchain)
(f) Explain the difference between contact and surface forces in materials science,
i.e. what is the molecular origin of each? Which one would you use a soft
cantilever (low k) and which one a stiff cantilever (high k)?
A. contact forces are described by the Herzian theory and originate from linear elastic
deformation of a body, i.e. compressing of covalent and noncovalent bonds within a
material, you would use a high k cantilever because contact forces are relatively strong
surface forces originate from noncovalent interactions, surface roughness and
mechanical interlocking, capillary forces due to water condensation, diffusion of polymers
chains across interfaces, etc. and result in adhesion, you would use a low k cantilever
because surface forces are relatively weak
1. The retraction data from an atomic force spectroscopy experiment is shown in the
figure below (*assume that D>>Rtip=15 nm and the Hamaker constant=10-19J) .
•
•
0
•
•
••• •
••• ••• • ••• • • • •
•
•
•
•
• •••• • III.
••
•
Force, F (nN)
•
I.
F~D-4
••
•
II.
0
Tip-Sample Separation
Distance, D (nm)
(a)
Q. In what region and how would the force curve change as one uses
increasingly softer cantilevers for this experiment, assuming that kcantilever still
remains<< ksample ? (*show schematically or describe in words)
A. Since the slope of the cantilever instability (region II) is equal to the cantilever
spring constant, the distance range of the cantilever instability and region of lost data
would be increased upon using softer cantilevers.
(b)
Q. At what distance has the interaction energy become comparable to the
thermal energy at room temperature ?
A. Let F=-C/D2 where C is a constant, then the net interaction energy is :
18
W(D)=-∫F(D)dD=-∫-C/D4dD=-C/D3~1/D3 (1)
If the atomic force spectroscopy experiment is modeled as a sphere interacting with
a surface for D>>Rtip:
WSPHERE-SFC(D)=-8π2Aρ2Rtip3/3(n-2)(n-3)Dn-3~1/Dn-3 (2)
Comparing eqs. (1) and (2), we find that the molecular interaction exponent, n=6.
Substituting n=6 into equation (2) we obtain:
WSPHERE-SFC(D)=-8π2Aρ2Rtip3/3(6-2)(6-3)D6-3=-2ΑRtip3/9D3 (3)
where : A is the Hamaker constant. Setting equation (3) equal to the thermal energy,
kBT :
B
W(D)=-2ΑRtip3/9D3 = -kBT (4)
Rearranging equation (4) to solve for D and substituting in appropriate numerical
values :
D=3√(2ARtip3)/(9•kBT)= 3√(10-19J•153 nm3)/(3•4.1•10-21J)= 26.3 nm
(2) Q. A carbon nanotube is attached to the end of the cantilever / probe tip to
which is attached a single positively charged protein molecule. A HRFS
experiment is conducted on a planar surface consists of freely rotating,
permanent dipoles. The cantilever/probe tip pulls off the surface at F=3 nN
and D=1 nm. How much of the force is due to purely electrostatic interactions
and how much is due to dispersion interactions (Adispersion=10-77 J m6, Acharge-freely-59
Jm4)?
rotating dipole=2.5 *10
A. The interaction potential between a molecule and a planar surface is :
WMOL-SFC(D)=-[2πAρ]/[(n-2)(n-3)Dn-3]
This is assuming an intermolecular pair potential of the following form:
w(r)=-A/rn
Substituting n=4 for the charge-freely rotating dipole interaction (page 28,
Israelachvili) into W(D) and taking the derivative with respect to D to obtain the
force one obtains :
W(D)charge-dipole=-[πAcharge-dipole ρ]/D
F(D)charge-dipole =-dW(D)/dD=-[πAcharge-dipoleρ]/D2 (1)
19
Substituting n=6 for the dispersion interaction (page 28, Israelachvili) into W(D)
and taking the derivative with respect to D to obtain the force one obtains :
W(D)dispersion=-[πAdispersionρ]/[6D3]
F(D)dispersion=-[πAdispersionρ]/[2D4] (2)
The ratio of the purely electrostatic to dispersion force is [equation (1) / equation
(2)]:
F(D)charge-dipole/F(D)dispersion = [Acharge-dipole/Adispersion]2D2
Substituting A values in and D=1 nm :
F(D)charge-dipole/F(D)dispersion=[2.5•10-59 Jm4•2•(1*1E-9m)2]/10-77 Jm6=5 (3)
The adhesion force is the sum of the dispersion and purely electrostatic interactions :
Fadhesion(D=1nm)= -3 nN= F(D)charge-dipole+F(D)dispersion (4)
Equations (3) and (4) give us two equations and two unknowns :
F(D)dispersion = -0.5 nN
F(D)charge-dipole = -4.5 nN
Energy (kBT)
3. Draw a schematic of a typical interatomic potential, U (kBT), versus separation
distance, r (nm), for an ionic bond (charge-charge) in water including the magnitude
of the energy and distance at the equilibrium bond length.
0
0
Separation Distance, r (nm)
20
A. Draw an asymmetric potential with EB≈5 kBT and re =0.25 nm.
4. Short Answer (*answers shouldn't be more than a few lines).
(a) Q. Would you expect the polymer film shown in the figure below to be
poly(methacrylic acid)(PMAA) or poly(acrylic acid) (PAA) and why?
n
n
CH3
O
OH
PMAA
O
OH
air
H2 0
po ly m er film
PAA
A. Since the H20 is shown to spread along or wet the polymer film, it has favorable
interactions with H20 and hence, is hydrophilic. Both polymers have hydrophilic
–-COOH groups which can form H-bonds, but the PMAA also has hydrophobic
methyl groups as well. Hence, one would it expect the polymer shown to be
PAA.
(b) Q. Draw a schematic of the interatomic potential between two nonpolar
atoms using a hard-sphere repulsion rather than a soft (power-law) repulsion.
A. w(r)=-A/rn+B/rm where n=6, m=∞
repulsion~rm, m⇒∞
w(r)
dispersion~r-6
r
(c) Q. Explain why poly(ethylene oxide), -[CH2-CH2-O]n- is hydrophilic and
soluble in water.
A. The polymer chain can H-bond via the oxygen (δ-) to the hydrogens (δ+) in H20.
(d) Q. What are the molecular origins of the two terms in the Lennard-Jones
potential?
A. attractive van der Waals-dispersion interactions and steric repulsion due to
overlap of electron clouds
21
(e) Q. In an AFM force spectroscopy experiment what happens at the
adhesion force ?
A. The force transducer (e.g. cantilever) suddenly detaches off the sample surface.
2. The data from an atomic force spectroscopy experiment conducted in aqueous salt
solution is shown in the figure below. The forces operating are dispersion van der Waals
forces and long-range electrostatic repulsive forces (*assume the Hamaker constant=10-19J,
Rtip=30 nm, D<<Rtip). You may neglect short-range, repulsive atomic steric forces.
(a) Calculate the Debye Length with a short explanation of your calculations included (you
may assume that electrostatic forces dominate the interaction for D>3 nm).
(b) For this sample, the solution salt concentration is varied so that the Debye length is 1 nm.
Calculate the magnitude of the force at D=0.5 nm.
(c) Using the Derjaguin approximations for dispersion interactions, calculate the magnitude
of the attractive dispersion component of the force if the experiment was conducted
between two spheres both with radius=5nm.
3
Force (nN)
2
1
0
-1
0
1
2
3
4
-2
-3
Distance (nm)
3. Short Answer Questions (please answer in one or a few sentences).
(a)
22
5
F1=Aexp(-κD1)
F2=Aexp(-κD2)
A. The following equation is for the DMT theory :
F(D)=AD3/2-Fadhesion
where : A=K√R
Fadhesion can be read directly off the graph and equals 1.83 nN. The DMT theory is expected
to hold for small D. Hence, choose one experimental datapoint for relatively small D to
determine A (e.g. F1=0 nN, D1=10 nm) :
23
F1=AD13/2-1.83
A=(F1+1.83)/D13/2
A=(0+1.83)/103/2
A=0.04 nN nm-3/2
calculate the magnitude of the repulsive component of the force at D=1 nm
(b) What is the magnitude of the repulsive force needed to make the system unstable
(purely attractive) at D=1 nm?
(c) *Using the Derjaguin approximations for dispersion interactions*, calculate the
magnitude of the attractive dispersion component of the force if the experiment was
conducted between two spheres both with radius=5nm.
A. (a) WSPHERE-SFC(D)= =-4π2Aρ2R/(n-2)(n-3)(n-4)(n-5)Dn-5 (D<<R)
for attractive dispersion interactions, n=6, and the Hamaker constant is A=Aπ2ρ2:
Wdispersion(D)=-ARtip/6D (D<<R)
Fdispersion (D)=-dW(D)/dD=-ARtip/6D2
Fdispersion(D=1 nm)=-(10-19 J•30•10-9m)/6(1•10-9m)2 =-0.5 nN
FTOTAL=Fdispersion+Frepulsive= -0.5 nN + Frepulsive=+2 nN (*read off from graph above)
Frepulsive=+2.5 nN
(b) FTOTAL=Fdispersion+Frepulsive= -0.5 nN + Frepulsive=0 nN
Frepulsive≤+0.5 nN
(c) The Derjaguin approximations for dispersion interactions give :
FSPHERE-SFC(D)=2πRWPLANES(D)
FSPHERE-SFC(D=1 nm)=-0.5 nN=2πRWPLANES(D=1nm)
WPLANES(D=1 nm)=-0.5 nN/2π(30 nm)=-2.65•10-3N/m
FSPHERE-SPHERE(D)=2π(R1R2/R1+R2)WPLANES(D)
R1=R2=R=5nm
24
FSPHERE-SPHERE(D=1 nm)=πRWPLANES(D=1nm)= π•5 nm•-2.65•10-3N/m=-0.0416 nN
This can be checked by :
Wdispersion(D)=-ARtip/12D (D<<R)
Fdispersion (D)=-dW(D)/dD=-ARtip/12D2
Fdispersion(D=1 nm)=-(10-19 J•5•10-9m)/12(1•10-9m)2 =-0.04166 nN
3. Short Answer Questions (*answers shouldn't be more than a few lines).
(a) Q. In an AFM experiment the adhesion force is measured to be 3 nN. If
the z-piezo deflection is 10 nm and the cantilever spring constant is 1 N/m
what is the tip-sample separation distance, D?
A. The cantilever behaves as a linear elastic Hookean spring (F=kδ)-hence, the
cantilever displacement is δ=F/k=-3 nN/1 N/m=-3nm.
The tip-sample separation distance is : D=z+δ=10 nm-3 nm=7nm
(b) Would you expect an attractive or repulsive force to be present between
two hydrophilic surfaces in aqueous solution?
A. repulsive, so that they can maximize their exposure to water (this is the
opposite of the hydrophobic attractive force)
(c) Would you expect hydrophobic forces to be present in organic solvents?
A. No! The origin of the attractive hydrophobic force is due to the avoidance of
water.
(d) The attractive van der Waals dispersion force ~α2 where : α is the
polarizability of the e- cloud. Explain this.
A. As the electron cloud is more easily distorted, the strength of the fluctuating
dipoles increases and hence, the attractive force increases.
(e) What is the difference between polar interactions and polarization
interactions?
A. Polar interactions are between an ion and a permanent dipole or two permanent
dipoles. Polarization interactions are between charge-nonpolar (induced or
instantaneous dipole) dipole-nonpolar (induced dipole).
25